## Base change for invariants [closed]

Possible Duplicate:
Invariants and base change

Suppose I have a commutative ring $R$, a free module $M$ of finite rank, and a finitely generated group $G$ acting on $M$ via $R$-linear endomorphisms.

Is there a nice condition on the action of $G$ on $M$ that would imply that taking invariants always commutes with base change, i.e. $$(M \otimes S)^{G} = M^G \otimes S$$ for any $R$-algebra $S$?

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 This is a duplicate of mathoverflow.net/questions/74621/…. Voted to close. – Ralph Feb 13 2012 at 1:27